Define a sphere: $S = \{ (x, y, z) \in \mathbb{R}^3 \big | (x - 1)^2 + y^2 + z^2 \leq 1 \}$ What is the triple integral of the scalar field $f(x, y, z)$ over $S$ in spherical coordinates? Assume that $x$, $y$, and $z$ are expressed in terms of $\rho$, $\theta$, and $\varphi$. Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^\pi \int_0^{2\pi} \int_{-1}^1 f(x - 1, y, z) \rho\sin^2(\varphi) \, d\rho \, d\theta \, d\varphi$ (Choice B) B $ \int_0^\pi \int_0^{2\pi} \int_0^1 f(x + 1, y, z) \rho^2\sin(\varphi) \, d\rho \, d\theta \, d\varphi$ (Choice C) C $ \int_0^\pi \int_0^{2\pi} \int_0^1 f(x - 1, y, z) \rho\sin^2(\varphi) \, d\rho \, d\theta \, d\varphi$ (Choice D) D $ \int_0^\pi \int_0^{2\pi} \int_{-1}^1 f(x + 1, y, z) \rho^2\sin(\varphi) \, d\rho \, d\theta \, d\varphi$
Explanation: The bounds we'll use are $0 < \theta < 2\pi$ and $0 < \varphi < \pi$. Here is the change of variables for spherical coordinates. $\begin{aligned} x &= \rho \cos(\theta) \sin(\varphi) \\ \\ y &= \rho \sin(\theta) \sin(\varphi) \\ \\ z &= \rho \cos(\varphi) \end{aligned}$ We want to represent the sphere $S$ with bounds in spherical coordinates. The standard unit sphere needs $\varphi$ to range across $[0, \pi]$, $\theta$ to range across $[0, 2\pi]$, and $\rho$ to range across $[0, 1]$. The sphere $S$ is a unit sphere because it has radius $1$, so our bounds look like this: $ \int_0^\pi \int_0^{2\pi} \int_0^1 \cdots \, d\rho \, d\theta \, d\varphi$ Although $S$ is a unit sphere, it's also shifted to the right by one unit on the $x$ -axis. How can we can keep the bounds for the unit sphere and integrate over $S$ ? We can add $1$ to $x$ everywhere to transform it one unit to the right. We write $f(x + 1, y, z)$ to account for this shift. This kind of a shift has no effect with change of variables, because its Jacobian is $1$. So our integral looks like this now: $ \int_0^\pi \int_0^{2\pi} \int_0^1 f(x + 1, y, z) \cdots \, d\rho \, d\theta \, d\varphi$ The final step is finding the Jacobian of spherical coordinates, which we'll need to multiply in to get the final integral. $J(\rho, \theta, \varphi) = \rho^2\sin(\varphi)$ [Derivation] The integral in spherical coordinates: $ \int_0^\pi \int_0^{2\pi} \int_0^1 f(x + 1, y, z) \rho^2\sin(\varphi) \, d\rho \, d\theta \, d\varphi$